1.

The one electron species having ionisation energy of 54.4 eV is

Answer»

`H`
`He^(+)`
`B^(4+)`
`Li^(2+)`

SOLUTION :Ionisation energy = `(13.6Z^(2))/(n^(2))eV=13.6Z^(2)`
for ONE electron species.
`therefore13.6Z^(2)=54.4` or, `Z^(2)=4` or `Z=2`
i.e., `He^(+)`.


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